Cannot display data from the database php

Question

I am working on a php project and I have a problem printing data from the database. The connection is working, the empty table is printed but when it comes to print the data, nothing is showed.My currently code:

 <?php


  $con = mysqli_connect('localhost','root','root', 'db'); 

  $user = $_SESSION['sess_user'];


  $query=("SELECT * FROM urls WHERE owner = '".$user."'");

  $qry_result = mysqli_query($con,$query);

  $display_string = "<table>";
  $display_string .= "<tr>";
  $display_string .= "<th> Id</th>";
  $display_string .= "<th> Url</th>";
  $display_string .= "<th> Owner </th>";
  $display_string .= "<th> SharingUrls </th>";
  $display_string .= "</tr>";

 while($row = mysqli_fetch_array($qry_result)) {//doesn't enter in this loop
    echo 'sdf';  //?? nothing
    $display_string .= "<tr>";
      $display_string .= "<td>$row[id]</td>";
      $display_string .= "<td>$row[url]</td>";
      $display_string .= "<td>$row[owner]</td>";
      $display_string .= "<td>$row[sharingUsers]</td>";

      $display_string .= "</tr>";
  }

  $display_string .= "</table>";

  echo $display_string;


?>

What can be the problem?

Where is session_start(); at your page??

Saty
– Saty

2016-06-10 07:34:02 +00:00
Commented Jun 10, 2016 at 7:34 — Saty
– Saty, Commented Jun 10, 2016 at 7:34
Notice : $_SESSION is undefined

Maninderpreet Singh
– Maninderpreet Singh

2016-06-10 07:40:32 +00:00
Commented Jun 10, 2016 at 7:40 — Maninderpreet Singh
– Maninderpreet Singh, Commented Jun 10, 2016 at 7:40

Praneeth · Accepted Answer · 2016-06-10 07:40:30Z

2

You are missing session_start()
That means your $user is not defined. That is the reason why there is no data fetched.

answered Jun 10, 2016 at 7:40

Praneeth

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1 Comment

Irina Farcau Over a year ago

This part of php was included in another php that's why I didn't write it here, but I had the session_start() included. The problem was in my db, owner field had the values 1,etc and my user was user1, that's why it didn't print anything. Thank you anyway

Megan Fox · Accepted Answer · 2016-06-10 07:42:09Z

0

Write session_start() at the top of your page to allow sessions to work on that page . Then try printing the value of echo $user = $_SESSION['sess_user'];

Once the value is got it should get the data from database . Alternately you can print the sql statement to check if the variable $user has the value.

echo $query=("SELECT * FROM urls WHERE owner = '".$user."'");

Hope this helps.

answered Jun 10, 2016 at 7:42

Megan Fox

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2 Comments

Irina Farcau Over a year ago

I wrote my solution in the first answer of the post. I gave you the correct answer because thanks to you I saw that it prints "user1" and then I realize that my owner field is different from the user and should be identically. Thanks

Megan Fox Over a year ago

Yeah that's the reason i asked you to print the data .

sanurah · Accepted Answer · 2016-06-10 07:49:33Z

First check if the $user value is set correctly. Then if it has any special characters you can use mysqli_real_escape_string. And in the table $row[id] should be $row['id']. I don't understand why you are creating a string. You can simply add the table in your html code with something like this.

<table>
    <thead>
    <tr>
        <th>id</th>
        <th>url</th>
        <th>owner</th>
        <th>sharingusers</th>
    </tr>
    </thead>
    <tbody>
    <?php
    if(isset($qry_result ) && !empty($qry_result )){
        while($row = $qry_result->fetch_assoc()){
            echo "
        <tr>
        <td>{$row['id']}</td>
        <td>{$row['url']}</td>
        <td>{$row['owner']}</td>
        <td>{$row['sharingusers']}</td>
        </tr>\n";
        }
    }
    ?>
    </tbody>
</table>

Passionate Coder · Accepted Answer · 2016-06-10 08:47:05Z

0

Hi You have not started session in this page. We have to start session_start at every page where we want to use session values. Session should be started at top of page before anything is displayed. Second please see your $user if it has value or not and add mysqli_error() to check if query has any error.so add this

$qry_result = mysqli_query($con,$query) or print_r(mysqli_error($con));

And please change this line

$display_string .= "<td>$row[id]</td>";

with $display_string .= "<td>".$row['id']."</td>";

this should be done in all other lines

answered Jun 10, 2016 at 8:47

Passionate Coder

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